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3 years ago

15

In ΔFGH, the measure of ∠H=90°, the measure of ∠F=52°, and FG = 4.3 feet. Find the length of HF to the nearest tenth of a foot.

2 answers:

swat323 years ago

6 0

Answer:

2.6

Step-by-step explanation:

EleoNora [17]3 years ago

5 0

Answer:2.6

Step-by-step explanation:

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A student said that since -9 is less than 4, then |-9| is less than |4|. Is the student correct? Explain why or why not.

IRISSAK [1]

No, that is not true. The absolute value is the distance a number is from zero. The distance of -9 to 0 is 9. 9 is not less than 4.

4 0

3 years ago

Read 2 more answers

Which equation has a constant of proportionality equal to 1? A. y=10/11x B. y= 7/8x C. y= 3/15x D. y=x

shtirl [24]

An equation is said to have a constant of proportionality if 'x' is a multiple of 'y' or 'y' is a multiple of 'x'.

Now, we have to determine an equation with constant of proportionality as '1'.

1. $y = \frac{10}{11}x$

This equation has constant of proportionality as ' $\frac{10}{11}$ '.

2. $y = \frac{7}{8}x$

This equation has constant of proportionality as ' $\frac{7}{8}$ '.

3. $y = \frac{3}{15}x$

This equation has constant of proportionality as ' $\frac{3}{15}$ '.

4. $y = x$

This equation has constant of proportionality as '1'.

Option D is the correct answer.

8 0

3 years ago

When lines the bottom of her first pan with aluminum foil. The area of the rectangle piece of foil is 11 1/4 in.². Its length is

ra1l [238]

Answer:

2.5 or 2 1/2 in

Step-by-step explanation:

Area = width x length

width = area / length = 11.25 / 4.5 = 2.5

7 0

3 years ago

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

How do i solve these?

never [62]

1/4

and 6/4 there we go here is your answer

6 0

4 years ago

Read 2 more answers